Let $U_r=cos(\theta +(r-1)a)$
and I have to find $f_{(r)}$ such that, $2sin(\frac{\alpha}{2})U_r=f_{(r+1)}-f_{(r)}$
I was managed to find $f_{(r)}$ such that $2sin(\frac{\alpha}{2})U_r=f_{(r+1)}-f_{(r-1)}$ where $f_{(r)}=sin(\theta +(r-1)a)$ and $\alpha=2a$
then $f_{(r+1)}-f_{(r-1)}=sin(\theta +(r)a)-sin(\theta +(r-2)a)\\=2cos(\frac{(\theta +(r)a)+(\theta +(r-2)a)}{2})sin(\frac{(\theta +(r)a)-(\theta +(r-2)a)}{2})\\=2sin(\frac{2a}{2})cos(\theta+(r-1)a)\\=2sin(\frac{2a}{2})U_r $
But unfortunately, how hard I tried I could not derive what they asked. I have a feeling that this problem might be wrong.
My problem is can anyone find $f_{(r)}$ such that $2sin(\frac{\alpha}{2})U_r=f_{(r+1)}-f_{(r)}$, does such $f_{(r)}$ exist?
Try using telescoping sum $$\sum_{r=1}^{n}(f_{r}-f_{r-1} )=f_n-f_0$$
Since you have $$f_{r}-f_{r-1} =2 \sin \left( \frac \alpha 2\right) \cos ( \theta +(r-2) \alpha) $$
$$\sum_{r=1}^{n}(f_{r}-f_{r-1}) =2 \sin \left( \frac \alpha 2\right) \sum_{r=1}^{n} \cos ( \theta +(r-2) \alpha)$$
Using the formula for sum of series of cosine, with angles in airthmatic progression, we can evaluate this sum on RHS.
We get $$f_n-f_0 = 2 \sin \alpha \cos \left( \theta + \frac{(n-3)}{2} \alpha \right) \left(\frac{\sin (n \alpha / 2)}{\sin ( \alpha / 2)} \right)$$
Let $f_0 =c$, which is an arbitrary constant.
Finally, your function is $$f_n = 2 \sin \alpha \cos \left( \theta + \frac{(n-3)}{2} \alpha \right) \left(\frac{\sin (n \alpha / 2)}{\sin ( \alpha / 2)} \right)+c$$